1 / 45

Tight Hardness Results for Some Approximation Problems [mostly Håstad]

Tight Hardness Results for Some Approximation Problems [mostly Håstad]. Adi Akavia Dana Moshkovitz S. Safra. “Road-Map” for Chapter I. Gap-3-SAT . expander. Gap-3-SAT-7 . Parallel repetition lemma. par[  , k] .  X  Y is NP-hard. Maximum Satisfaction. Def : Max-SAT Instance :

alaura
Download Presentation

Tight Hardness Results for Some Approximation Problems [mostly Håstad]

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Tight Hardness Results forSome Approximation Problems[mostly Håstad] Adi AkaviaDana MoshkovitzS. Safra

  2. “Road-Map” for Chapter I Gap-3-SAT  expander Gap-3-SAT-7  Parallel repetition lemma par[, k]  XY isNP-hard

  3. Maximum Satisfaction Def: Max-SAT • Instance: • A set of variables Y = { Y1, …, Ym } • A set of Boolean-functions (local-tests) over Y = { 1, …, l } • Maximization: • We define () = maximum, over all assignments to Y, of the fraction of i satisfied • Structure: • Various versions of SAT would impose structure properties on Y, Y’s range and 

  4. Max-E3-Lin-2 Def: Max-E3-Lin-2 • Instance: a system of linear equationsL = { E1, …, En } over Z2each equation of exactly 3 variables(whose sum is required to equal either 0 or 1) • Problem: Compute (L)

  5. 2-Variables Functional SAT Def[ XY ]:  over • variables X,Y of range Rx,Ry respectively • each  is of the form xy: RxRy an assignment A(xRx,yRy)satisfies xyiffxy(A(x))=A(y) [ Namely, every value to X determines exactly 1 satisfying value for Y] Thm: distinguishing between •  A satisfies all  •  A satisfies < fraction of  IS NP-HARD as long as |Rx|,|Ry|> -0(1)

  6. Proof Outline Gap-3-SAT ? Gap-3-SAT-7 Def: 3SAT is SAT where every iisa disjunction of 3 literals. Def: gap-3SAT-7 is gap-3SAT with theadditional restriction, that everyvariable appears in exactly 7 local-tests Theorem: gap-3SAT-7 is NP-hard par[, k] XY isNP-hard

  7. Expanders Gap-3-SAT ? Expanders Gap-3-SAT-7 Def: a graph G(V,E) is a c-expander if for everySV, |S| ½|V|: |N(S)\S|  c·|S| [where N(S) denotes the set of neighbors of S] Lemma: For every m, one can construct in poly-time a 3-regular, m-vertices, c-expander, for some constant c>0 Corollary: a cut between S and V\S, for |S|  ½|V| must contain > c·|S| edges par[, k] XY isNP-hard

  8. Reduction Using Expanders Assume ’ for which (’) is either 1 or 1-20/c.  is ’ with the following changes: • an occurrence of y in i is replaced by a variable xy,i • Let Gy, for every y, be a 3-regular, c-expander over all occurrences xy,iof y • For every edge connecting xy,i to xy,j in Gy, add to  the clauses (xy,i  xy,j) and (xy,i   xy,j) It is easy to see that: • ||  10 |’| • Each variable xy,iof appears in exactly 7i   constructible by the Lemma ensuring equality

  9. Correctness of the Reduction •  is completely satisfiable iff ’ is • In case ’ is unsatisfiable: (’) <1-20/cLet A be an optimal assignment to Let Amajassign xy,i the value assigned by A to the majority, over j, of variables xy,jLet FA and FAmaj be the sets of   unsatisfied by A and Amaj respectively: ||·(1-()) = |FA| = |FAFAmaj|+|FA\FAmaj| |FAFAmaj|+½c|FAmaj\FA|  ½c|FAmaj| andsince Amaj is in fact an assignment to ’()  1- ½c(1- (’))/10 < 1- ½c(20/c)/10= 1-

  10. Notations Def: For a 3SAT formula  over Boolean variables Z, • Let Zk be the set of all k-sequences of ’s variables • Let k be the set of all k-sequences of ’s clauses Def: For any VYk and Ck, let • RYbe the set of all assignments to V • RXbe the set of all satisfying assignments to C Def: For any set of k variables VZk, and a set of k clauses Ck, denote V CV is a choice of one variable of each clause in C.

  11. Parallel SAT Def: for a 3SAT formula  over Boolean variables Z, define par[, k]: par[, k]has two types of variables: • yV for every set VYk,where yV‘s range is the set RYof all assignments to V • xC for every set Ck,where xC‘s range is the set RXof all satisfying assignments to all clauses in C par[, k] has a local-test [C,V]for each V Cwhich accepts if xC’s value restricted to V is yV’s value (namely, if the assignments to T[C] and T[V] are consistent) |RY|=2k |RX|=7k

  12. Gap Increases with k Gap-3-SAT Gap-3-SAT-7 Parallel repetition lemma par[, k] Note that if () = 1 then (par[, k]) = 1 On the other hand, if  is not satisfiable: Lemma: (par[, k])  ()c·kfor some c>0 Proof: first note that1-(par[, 1])  (1-())/3 now, to prove the lemma, apply the Parallel-Repetition lemma [Raz] to par[, 1] XY isNP-hard In any assignment tos variables, any unsatisfied clause in”induces“ at least 1 (out of corresponding 3) unsatisfied par[, 1]

  13. Conclusion: XY isNP-hard • Denote: •  = par[, k] • X={xC} • Y={yV} Then, • distinguish between: •  A satisfies all  •  A satisfies < fraction of  IS NP-HARD as long as |Rx|,|Ry|> -0(1)

  14. “Road-Map” for Chapter II XY isNP-hard Long code L  () = ()  LLC-Lemma: (L) = ½+/2  (par[,k]) > 42

  15. Main Theorem Thm: gap-Max-E3-Lin-2(1-, ½+)is NP-hard. That is, for every constant 0<<¼ it is NP-hard to distinguish between the case where 1- of the equations are satisfiable and the case where ½+ are. [ It is therefore NP-Hard to approximateMax-E3-Lin-2 to within factor 2- for any constant 0<<¼]

  16. This bound is tight • A random assignment satisfies half of the equations. • Deciding whether a set of linear equations have a common solution is in P (Gaussian elimination).

  17. XY isNP-hard Long code L  LLC-Lemma: (L) = ½+/2  (par[,k]) > 42 Distributional Assignments Let  be a SAT instance over variablesZ of range R. Let (R) be all distributions over R Def: a distributional-assignment to  is A: Z (R)Denote by () themaximum over distributional-assignments A of theaverage probability for  to be satisfied,if variables` values are chosen according to A Clearly()  (). Moreover Prop: ()  ()

  18. Distributional-assignment to  1 1 0 1 0 0 0 1 x1 x3 x2 xn OR: 1 0 1 1 0 0 1 0 x1 x3 x2 xn

  19. Restriction and Extension Def: For any yYover RY and xXover RXs.t xy • The natural restrictionof an aRXto Ry is denoted a|y • The elevationof a subset FP[RY] to RX is the subset F*P[RX] of all members a of RX for which xy(a) FF* = { a | a|y F }

  20. XY isNP-hard Long-Code Long code L In the long-code the set of legal-words consists of all monotone dictatorships This is the most extensive binary code, as its bits represent all possible binary values over n elements LLC-Lemma: (L) = ½+/2  (par[,k]) > 42

  21. Long-Code • Encoding an element e[n]: • Eelegally-encodes an element e if Ee = fe T F F T T

  22. Long-Code over Range R |BP[R]| = 2|R|-1-1 BP[R]  the set of all subsets of R of size ≤½|R| • Our long-code: in our context there’re two types of domains “R”:Rx and Ry . Def: an R-long-code has 1 bit for each F P[R] namely, any Boolean function: P[R]  {-1, 1} Def: a legal-long-code-word of an element eR, is a long-code ERe: P[R]  {-1, 1} that assigns eF to every subset F  P[R]

  23. Linearity of a Legal-Encoding An assignment A : BP[R]  {-1,1}, if legal, is a linear-function, i.e.,  F, G  BP[R]:f(F)  f(G)  f(FG) Unfortunately, any character is linear as well!

  24. XY isNP-hard The Variables of L Long code L Consider  (xy) forlarge constant k (to befixed later) L has 2 types of variables: • a variable z[y,F] for every variable yY and a subset F  BP[Ry] • a variable z[x,F] for every variable xX and a subset F  BP[Rx] LLC-Lemma: (L) = ½+/2  (par[,k]) > 42

  25. The Distribution  Def: denote by  the distribution over all subset of Rx, which assigns probability to a subset H as follows:Independently, for each a  Rx, let • aH with probability 1- • aH with probability  One should think ofas a multiset of subsets in which every subsetHappears with the appropriate probability

  26. Linear equation L‘s multiplicative-equations are the union, over all xy  , of the following:FP[RY], GP[RX] and H(RX)z(y,F) z(x, G) = z(x, F*GH)

  27. Multiplicative representation General Fourier Analysis facts Revised Representation Representation by Fourier Basis Multiplicative Representation: • True  -1 • False  1 • L: • z[X,*], z[Y,*]  {-1, 1} • z[X,f] • z[Y, g] • z[X,’f•g•h’] = 1 Claim 2 Claim 1 Claim 3:The expected success of the distributional assignment  on [C,V]par[,k] is at least 4 2 (par[,k]) > 42 

  28. XY isNP-hard L  Prop: if () = 1 then (L) = 1- Proof:Let A be a satisfying assignment to . Assign all variables of Laccording to the legal encoding of A’s values.A linear equation of L, corresponding to X,Y,F,G,H, would be unsatisfied exactly if xH, which occurs with probability  over the choice of H. LLC-Lemma: (L) = ½+/2() > 42 LLC-Lemma: (L) = ½+/2  (par[,k]) > 42 Note: independent of k! (Later we use that fact to define k large enough for our needs).  = 2(L) -1

  29. Hardness of approximating Max-E3-Lin-2 Main Theorem: For any constant >0: gap-Max-E3-Lin-2(1-,½+) is NP-hard. Proof: Let’ be a gap-3SAT-7(1, 1-)By proposition(’) = 1 (L’)  1- 

  30. Lemma  Main Theorem Prop:Let  be a constant >0 s.t.:(1-)/(½+/2)  2-Let k be large enough s.t.:43 > (’)c·kThen (’) < 1 (L’)  ½+/2  ½+  Proof: Assume, by way of contradiction, that (L)  ½+/2 then:43 > (’)c·k  () > 42, which implies that  > .Contradiction! of the parallel repetition lemma

  31. Long-Code as an inner product space Def: { A : BP[R]  {-1,1} } is an inner-product space: A,B{ A : BP[R]  {-1,1} }

  32. An Assignment  to L For any variable xXThe set z[x,*] of variables ofL represent the long-code of xLet be the Fourier-Coefficient <|z[X,*],s> For any variable yYThe set z[y,*] of variables ofL represent the long-code of yLetbe the Fourier-Coefficient <|z[Y,*],s>

  33. The Distributional Assignment. Def: Let  be a distributional-assignmentto  as follows: • For any variable x • Choose a set SRx with probability , • Uniformly choose a random assignment aS. • For any variable y • Choose a set SRy with probability , • Uniformly choose a random assignment bS.

  34. Longcode and Fourier Coeficients go to claim2 Auxiliary Lemmas: 1. For any F,GBP[R] and S  R, S(F·G) = S(F)·S(G). 2. For any FBP[R] and s,s’  R, s(F)·s’(F) = ss’(F) 3. For any random F (uniformly chosen) and S, E[ s(F) ]=0 and E[ (F) ]=1. =xf(x) apply multiplication’s commutative & associative properties (f)·(f)=xf(x)·xf(x)= xf(x)2·x(f)=1·x(f) xs, f(x)is 1or-1with probability ½

  35. Home Assignment • Given an assignment to a Longcode A:BP[R]  {-1, 1}, show that for any (constant)  > 0, there is a constant h(), which depends on , however does not depend on R such that:| {e  R | (Ee, A) > ½ +  } |  h()where (A1, A2)is the fraction of bits A1 and A2differ on.

  36. What’s Ahead: We show ‘s expected success on xy is > 42 in two steps: First we show (claim 1) that ‘s success probability, for any xy is Then show (claim 3) that value to be 42

  37. General Fourier Analysis facts Claim 1 Multiplicative representation go to claim3 Representation by Fourier Basis Claim 1 Claim 1: The success probability of  on xy  is Proof:That success probability is at least and if S’=S|y there is at least one bS s.t. b|y  S’ So, ‘s success probability is at least |S|-1times the case in which the chosen S’ and S satisfy: S|y = S’, i.e. at least Claim 2 Claim 3:The expected success of the distributional assignment  on [C,V]par[,k] is at least 4 2 (par[,k]) > 42 

  38. Lemma’s Proof - Claim 2 (1) go to claim3 General Fourier Analysis facts Multiplicative representation Claim 2: Proof: The test accepts iffz[y,F]•z[x, G]•z[x,F*•G•H] = 1 By our assumption, this happens with probability/2+½. Now, according to the definition of the expectation: Exy, F, G, H[z[y,F]•z[x, G]•z[x,F*•G•H]] = 1•(½+/2) + (-1)•(1 -(½+/2)) =  Representation by Fourier Basis Claim 2 Claim 1 Claim 3:The expected success of the distributional assignment  on [C,V]par[,k] is at least 4 2 (par[,k]) > 42 

  39. Lemma’s Proof - Claim 2 (2) We next show that Hence,

  40. Lemma’s Proof - Proposition

  41. Lemma’s Proof - Claim 3 Claim 3: The expected success of the distributional assignment on xy  is at least 42 Proof:Claim 1 gives us the initial lower bound for the expected success:

  42. Lemma’s Proof - Claim 3 As we’ve already seen, . Hence, our lower-bound takes the form of Or alternatively, Which allows us to use the known inequality E[x2]E[x]2 and get

  43. Lemma’s Proof - Claim 3 Byauxiliary lemmas(4||)-1/2  e-2||  (1-2)||, i.e.||-1/2 (4)1/2 ·(1-2)||, which yields the following bound That is, Now applying claim 2 results the desired lower bound

  44. Lemma’s Proof -Conclusion We showed that there is an assignment scheme with expected success of at least 42 ,  There exists an assignment that satisfies at least 42 of the tests in   () > 42 Q.E.D.

  45. Home Assignment Show it is NP-hard, for any  > 0, given a 3SAT instance , to distinguish between the case where () = 1, and the case in which () < 7/8+ Hint: Let ’s variables be as in L, and ’s clauses to take the formF OR G OR F*GHfor f and g chosen in the same way as in L,while h is chosen as follows: • H(b) = 1 for b such that F(b|V) and G(b) are both FALSE • For all other b’s, independently for each b, H(b)=1 with probability , and -1 with probability 1-

More Related